Metamath Proof Explorer


Theorem clatlubcl

Description: Any subset of the base set has an LUB in a complete lattice. (Contributed by NM, 14-Sep-2011)

Ref Expression
Hypotheses clatlubcl.b
|- B = ( Base ` K )
clatlubcl.u
|- U = ( lub ` K )
Assertion clatlubcl
|- ( ( K e. CLat /\ S C_ B ) -> ( U ` S ) e. B )

Proof

Step Hyp Ref Expression
1 clatlubcl.b
 |-  B = ( Base ` K )
2 clatlubcl.u
 |-  U = ( lub ` K )
3 eqid
 |-  ( glb ` K ) = ( glb ` K )
4 1 2 3 clatlem
 |-  ( ( K e. CLat /\ S C_ B ) -> ( ( U ` S ) e. B /\ ( ( glb ` K ) ` S ) e. B ) )
5 4 simpld
 |-  ( ( K e. CLat /\ S C_ B ) -> ( U ` S ) e. B )